# The equation of the line through the point (0,1,2)

Question:

The equation of the line through the point $(0,1,2)$ and perpendicular to the line

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is :

1. $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}$

2. $\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}$

3. $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}$

4. $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

Correct Option: , 4

Solution:

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

$\Rightarrow \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2 \mathrm{r}+1,3 \mathrm{r}-1,-2 \mathrm{r}+1)$

Since, $\overline{\mathrm{QP}} \perp(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$

$\Rightarrow 4 r+2+9 r-6+4 r+2=0$

$\Rightarrow r=\frac{2}{17}$

$\Rightarrow P\left(\frac{21}{17}, \frac{-11}{17}, \frac{13}{17}\right)$

$\Rightarrow \overline{\mathrm{PQ}}=\frac{21 \hat{\mathrm{i}}-28 \hat{\mathrm{j}}-21 \hat{\mathrm{k}}}{17}$

So, $\overline{\mathrm{QP}}: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-1}{4}=\frac{\mathrm{z}-2}{3}$